Question

116. 填充每个节点的下一个右侧节点指针

English Version

题目描述

给定一个 完美二叉树 ，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。

初始状态下，所有 next 指针都被设置为 NULL。

 

示例 1：

输入：root = [1,2,3,4,5,6,7]
输出：[1,#,2,3,#,4,5,6,7,#]
解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。序列化的输出按层序遍历排列，同一层节点由 next 指针连接，'#' 标志着每一层的结束。

示例 2:

输入：root = []
输出：[]

 

提示：

• 树中节点的数量在 [0, 212 - 1] 范围内
• -1000 <= node.val <= 1000

 

进阶：

• 你只能使用常量级额外空间。
• 使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。

解法

方法一：BFS

使用队列进行层序遍历，每次遍历一层时，将当前层的节点按顺序连接起来。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点个数。

"""
# Definition for a Node.
class Node:
  def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
    self.val = val
    self.left = left
    self.right = right
    self.next = next
"""


class Solution:
  def connect(self, root: "Optional[Node]") -> "Optional[Node]":
    if root is None:
      return root
    q = deque([root])
    while q:
      p = None
      for _ in range(len(q)):
        node = q.popleft()
        if p:
          p.next = node
        p = node
        if node.left:
          q.append(node.left)
        if node.right:
          q.append(node.right)
    return root

/*
// Definition for a Node.
class Node {
  public int val;
  public Node left;
  public Node right;
  public Node next;

  public Node() {}

  public Node(int _val) {
    val = _val;
  }

  public Node(int _val, Node _left, Node _right, Node _next) {
    val = _val;
    left = _left;
    right = _right;
    next = _next;
  }
};
*/

class Solution {
  public Node connect(Node root) {
    if (root == null) {
      return root;
    }
    Deque<Node> q = new ArrayDeque<>();
    q.offer(root);
    while (!q.isEmpty()) {
      Node p = null;
      for (int n = q.size(); n > 0; --n) {
        Node node = q.poll();
        if (p != null) {
          p.next = node;
        }
        p = node;
        if (node.left != null) {
          q.offer(node.left);
        }
        if (node.right != null) {
          q.offer(node.right);
        }
      }
    }
    return root;
  }
}

/*
// Definition for a Node.
class Node {
public:
  int val;
  Node* left;
  Node* right;
  Node* next;

  Node() : val(0), left(NULL), right(NULL), next(NULL) {}

  Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

  Node(int _val, Node* _left, Node* _right, Node* _next)
    : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
  Node* connect(Node* root) {
    if (!root) {
      return root;
    }
    queue<Node*> q{{root}};
    while (!q.empty()) {
      Node* p = nullptr;
      for (int n = q.size(); n; --n) {
        Node* node = q.front();
        q.pop();
        if (p) {
          p->next = node;
        }
        p = node;
        if (node->left) {
          q.push(node->left);
        }
        if (node->right) {
          q.push(node->right);
        }
      }
    }
    return root;
  }
};

/**
 * Definition for a Node.
 * type Node struct {
 *   Val int
 *   Left *Node
 *   Right *Node
 *   Next *Node
 * }
 */

func connect(root *Node) *Node {
  if root == nil {
    return root
  }
  q := []*Node{root}
  for len(q) > 0 {
    var p *Node
    for n := len(q); n > 0; n-- {
      node := q[0]
      q = q[1:]
      if p != nil {
        p.Next = node
      }
      p = node
      if node.Left != nil {
        q = append(q, node.Left)
      }
      if node.Right != nil {
        q = append(q, node.Right)
      }
    }
  }
  return root
}

/**
 * Definition for Node.
 * class Node {
 *   val: number
 *   left: Node | null
 *   right: Node | null
 *   next: Node | null
 *   constructor(val?: number, left?: Node, right?: Node, next?: Node) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *     this.next = (next===undefined ? null : next)
 *   }
 * }
 */

function connect(root: Node | null): Node | null {
  if (root == null || root.left == null) {
    return root;
  }
  const { left, right, next } = root;
  left.next = right;
  if (next != null) {
    right.next = next.left;
  }
  connect(left);
  connect(right);
  return root;
}

方法二：DFS

使用递归进行前序遍历，每次遍历到一个节点时，将其左右子节点按顺序连接起来。

具体地，我们设计一个函数 $dfs(left, right)$，表示将 $left$ 节点的 $next$ 指针指向 $right$ 节点。在函数中，我们首先判断 $left$ 和 $right$ 是否为空，若都不为空，则将 $left.next$ 指向 $right$，然后递归地调用 $dfs(left.left, left.right)$, $dfs(left.right, right.left)$, $dfs(right.left, right.right)$。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点个数。

"""
# Definition for a Node.
class Node:
  def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
    self.val = val
    self.left = left
    self.right = right
    self.next = next
"""


class Solution:
  def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
    def dfs(left, right):
      if left is None or right is None:
        return
      left.next = right
      dfs(left.left, left.right)
      dfs(left.right, right.left)
      dfs(right.left, right.right)

    if root:
      dfs(root.left, root.right)
    return root

/*
// Definition for a Node.
class Node {
  public int val;
  public Node left;
  public Node right;
  public Node next;

  public Node() {}

  public Node(int _val) {
    val = _val;
  }

  public Node(int _val, Node _left, Node _right, Node _next) {
    val = _val;
    left = _left;
    right = _right;
    next = _next;
  }
};
*/

class Solution {
  public Node connect(Node root) {
    if (root != null) {
      dfs(root.left, root.right);
    }
    return root;
  }

  private void dfs(Node left, Node right) {
    if (left == null || right == null) {
      return;
    }
    left.next = right;
    dfs(left.left, left.right);
    dfs(left.right, right.left);
    dfs(right.left, right.right);
  }
}

/*
// Definition for a Node.
class Node {
public:
  int val;
  Node* left;
  Node* right;
  Node* next;

  Node() : val(0), left(NULL), right(NULL), next(NULL) {}

  Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

  Node(int _val, Node* _left, Node* _right, Node* _next)
    : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
  Node* connect(Node* root) {
    function<void(Node*, Node*)> dfs = [&](Node* left, Node* right) {
      if (!left || !right) {
        return;
      }
      left->next = right;
      dfs(left->left, left->right);
      dfs(left->right, right->left);
      dfs(right->left, right->right);
    };
    if (root) {
      dfs(root->left, root->right);
    }
    return root;
  }
};

/**
 * Definition for a Node.
 * type Node struct {
 *   Val int
 *   Left *Node
 *   Right *Node
 *   Next *Node
 * }
 */

func connect(root *Node) *Node {
  var dfs func(*Node, *Node)
  dfs = func(left, right *Node) {
    if left == nil || right == nil {
      return
    }
    left.Next = right
    dfs(left.Left, left.Right)
    dfs(left.Right, right.Left)
    dfs(right.Left, right.Right)
  }
  if root != nil {
    dfs(root.Left, root.Right)
  }
  return root
}

/**
 * Definition for Node.
 * class Node {
 *   val: number
 *   left: Node | null
 *   right: Node | null
 *   next: Node | null
 *   constructor(val?: number, left?: Node, right?: Node, next?: Node) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *     this.next = (next===undefined ? null : next)
 *   }
 * }
 */

function connect(root: Node | null): Node | null {
  if (root == null) {
    return root;
  }
  const queue = [root];
  while (queue.length !== 0) {
    const n = queue.length;
    let pre = null;
    for (let i = 0; i < n; i++) {
      const node = queue.shift();
      node.next = pre;
      pre = node;
      const { left, right } = node;
      left && queue.push(right, left);
    }
  }
  return root;
}